Mathematical Tool of Quantum Mechanics | Quantum Science Philippines

## Mathematical Tool of Quantum Mechanics

#### Problem:

Consider the set of all entities of the form $(a, b, c)$ where the entries are real numbers. Addition and scalar multiplication are defined as follows:

$\begin{array}{ccc} (a, b, c) + (d, e, f) &=& (a+d, b+e, c+f)\\ a(a, b, c) &=& (aa, ab,ac) \end{array}$

i.) Write down the null vector and inverse of $(a, b, c)$.

ii.) Show that vectors of the form $(a, b, 1)$ do not form a vector space.

#### Solution:

i.) Let $(x, y, z)$ be the null vector of $(a, b, c)$, such that

$(a, b, c) + (x, y, z) = (a, b, c)$

Then as defined,

$(a+x, b+y, c+z) = (a, b, c)$

and by comparison,

$\begin{array}{ccc} a+x &=& a \\ b+y &=& b \\ c+z &=& c \end{array}$

If we are to solve these equations for the values of $x, y$ and $z$, we could get

$\begin{array}{ccc} x &=& 0 \\ y &=& 0 \\ z &=& 0 \end{array}$.

Therefore, the null vector is given as

$\mathbf{(x, y, z)} = \mathbf{(0, 0, 0)}$.

Now, we let the inverse of  $(a, b, c)$ be  $(\alpha, \beta, \gamma)$, such that following the definition of an inverse vector we have

$(a, b, c) + ( \alpha , \beta , \gamma) = (0, 0, 0)$

$( a + \alpha , b + \beta , c+ \gamma ) = (0, 0, 0)$

Following the same process as before, if we compare we can obtain

$\begin{array}{ccc} a+\alpha &=& 0 \\ b+\beta &=& 0 \\ c+\gamma &=& 0 \end{array}$

Solving for $\alpha, \beta$ and $\gamma$,

$\begin{array}{ccc} \alpha &=& -a \\ \beta &=& -b \\ \gamma &=& -c \end{array}$.

Therefore, the inverse vector of $(a, b, c)$ is

$\mathbf{( \alpha, \beta, \gamma )} = \mathbf{( -a, -b, -c)}$.

ii.) The vector  $(a, b, 1)$ do not form a vector space because:

a.) it violates the closure under addition,

$\begin{array}{ccc} ( a, b, 1) + ( c, d, 1) &=& ( a+c, b+d, 1+1) \\ &=& \mathbf{( a+c, b+d, 2) \not\in (a,b, 1)} \end{array}$.

b.) it violates the closure under scalar multiplication,

$\lambda (a, b, 1) = \mathbf{( \lambda a, \lambda b, \lambda ) \not\in ( a, b, 1)}$.

c.) there is NO null vector

$(0, 0, 0) \not\in (a, b, 1)$

d. an additive inverse does not exist

$( -a, -b, -1) \not\in (a, b, 1)$.